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kobusy [5.1K]
1 year ago
14

A blue train of mass 50 kg moves at 4 m/s toward a green train of 30 kg initially at rest. What is the initial momentum of the b

lue and green train combined?
A. 20 kgm/s
B. 50 kgm/s
C. 0 kgm/s
D. 200 kgm/s
Physics
1 answer:
emmainna [20.7K]1 year ago
4 0

<em>The correct option is</em> D.   The initial momentum of the blue and green train combined during the collision is 200 kgm/s.

<h3>Initial momentum of the blue and green train</h3>

Apply the principle of conservation of linear momentum as follows;

Pi = m1v1 + m2v2

where;

  • m1 is mass of blue train
  • m2 is mass of green train
  • v1 is velocity of blue train
  • v2 is velocity green train
  • Pi is the initial momentum of the two trains

Pi = (50 x 4) + 30(0)

Pi = 200 kgm/s

Thus, the initial momentum of the blue and green train combined is 200 kgm/s.

Learn more about momentum here: brainly.com/question/7538238

#SPJ1

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Answer:

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Explanation:

For point a:

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W=0\\\\Q=0\\\\m_e=0

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Interpolate the enthalpy between T = 300 \ K \ and\ T=295\ K. The surrounding air  

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T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

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Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

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Answer:

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